Analog Circuits

Day-6
 BJT Amplifiers
Introduction
• In this unit we learn about AC response of transistors using different transistor
models.
• In AC analysis we have to decide whether to use small signal or large signal
technique. In this unit we will use small signal as input.
• Large signal amplifiers are power amplifiers.
Operating Point in Small Signal Analysis:
• In small signal analysis, as the input signal variation is small, the output signal
variation is also limited and hence swing in Q-point is also limited.
Small signal is defined as the signal having magnitude sufficiently small to keep
transistor in active region.
Transistor as an Amplifier
To use transistor as an amplifier it must be in the active region.

We know that Xc=1/2*pi*f*c

For AC input f≠0(f not equal to zero)
Xc=1/large value=0(approximately)
Hence for AC input the capacitors are
short circuited.
• C1,C2 are coupling capacitors because
those are the capacitors which are
coupling the input to the amplifier
and output to the load respectively.
Why to use coupling capacitors?
In order to prevent the DC from previous stage to interfere with the Vcc and hence the
operating point will remain constant.
C3 is bypass capacitor because it bypasses the AC signal as RE offers some resistance.
To find out AC response we need to do 2 things
1)Obtain AC equivalent circuit.
2)Replace the transistor with equivalent circuit.
Equivalent circuit for AC analysis:
STEP-1
Short circuit the DC sources.
In the figure we have 1 DC source (ie) Vcc
and the potential of ground is 0v.
STEP-2
Short all the capacitors C1,C2 and C3
STEP-3
Redraw the network removing all the
elements which are short circuited in STEP-
1 and STEP-2.
 Problems
1) If the emitter resistance in a common-emitter voltage amplifier is
not bypassed, it will GATE-2014
A) Reduces voltage gain and input impedance
B) Reduces voltage gain and increase the input impedance
C) Increases voltage gain and reduces the input impedance
D) Increases both voltage gain and the input impedance
Solution: B
Reduces voltage gain and increases input impedance due to
feedback.
The equivalent model of transistor:

Equivalent model is the combination of circuit elements properly chosen to best

represent the actual behaviour of the device under specific Operating Point.
We need equivalent models to use these network theorems in order to find out
different network parameters like in case of P-N diode.
There are three equivalent models of transistors. They are
1)Hybrid model (for low frequencies)
2)re –model (for low and high frequencies)
3)Hybrid ∏-model (for low and high frequencies)
All the three models are used for small signal analysis.
Hybrid model
It is also known as h-parameters model

1) It is widely used before the popularity of re-model.

2) Parameters are defined in general terms for any operating point conditions.
In hybrid model we have to calculate h-parameters and using them we will draw the
equivalent circuit.
• These parameters have mixed dimensions hence these are known as hybrid
parameters.
Need for Hybrid parameters:
Before transistors, vacuum tubes are used to design circuits. We have four parameters
in small signal amplifiers. All these are obtained only by z-parameters or by y-
parameters in case of vacuum tubes. In case of transistors there was problem
determining z-parameters, so a new set of parameters called as hybrid parameters are
introduced.
 Port 1 current is I1 Potential difference across port-1 is V1
Port 2 current is I2 Potential difference across port-2 is V2

Total current or voltage value=ac value + dc value

We can define parameters by taking any 2 parameters out of 4 as dependent and rest
2 as independent.
Let V1 and I2 be dependent quantities.
I1 and V2 are independent quantities.
Say V1 and I2 are functions of I1 and V2
V1=f1(I1,V2); I2=f2(I1,V2)
v1=h11*i1+h12*v2--------------------------(1)
i2=h21*i1+h22*v2---------------------------(2)
These equations are applicable to all 3 transistor configurations(CC,CB,CE)
If we substitute v2=0 in Eq-1 and Eq-2, then
h11=v1/i1 ; where h11 is input impedance when output is short circuited
Hence h11 can be represented as hi
h21= i2/i1 ; where h21 is forward current gain when the output is short circuited
Hence h21 can be represented as hf
If we substitute i1=0 in Eq-1 and Eq-2, then
h12=v1/v2 ; where h12 is reverse voltage gain when input is open circuited
Hence h12 can be represented as hr
h22= i2/v2 ; where h22 is admittance with input open circuited
Hence h22 can be represented as ho
Nomenclature of h-parameters for various transistor configurations
We can add ‘e’ or ‘b’ or ‘c’ as second suffix to all h-parameters for CE, CB and CC
configurations respectively
Example: For CE, the h-parameters are
hihie hfhfe hrhre hohoe
To draw equivalent circuit for Eq-1 and Eq-2
Let us consider Eq-1,
v1=h11*i1+h12*v2
Unit of each term in the equation is volts. Hence by applying KVL we can obtain
equivalent circuit,
v1 = hi*i1+hr*v2
Let us consider Eq-2,
i2=h21*i1+h22*v2
Use KCL to obtain equivalent circuit,
i2=hf*i1+ho*v2
 ==

Note: Whenever we have the transistor in the circuit and you have to perform
AC analysis then the conventional transistor symbol is replaced by equivalent
model. If we want to make the equivalent model according to transistor
configuration, an other subscript will be added.
Let CE be the transistor, hihie; hfhfe; hrhre; hohoe
Analysis of transistor amplifier using h-parameter

In the equivalent circuit of the transistor, introduce source voltage Vs and resistance
Rs on the input side.
On the output side, introduce a load resistance RL and iL is current through load
resistance.
Zo-output impedance; Zi-input impedance
Expression for current gain:
Current gain is defined as the ratio of output current to input current. It is denoted by
Ai. Ai=iL/i1
iL=-i2 (from above figure)
Voltage drop across RL is V2
V2=iL*RL=-i2*RL
Consider the equation from equivalent model of transistor
i2=hf*i1+ho*v2 hf*i1+ho*(-i2*RL)
By solving the above equation we get, i2/i1=hf/(1+ho*RL)
Therefore, Ai = iL/i1 =-hf/(1+ho*RL)
This is true for all transistor configurations.
Expression for Input impedance:
Input impedance is defined as the impedance seen from terminals 1 and 11
According to ohms law, V1=i1*Zi
𝑉1
Zi=
𝑖1

Consider Eq-1, v1=hi*i1+hr*v2= hi*i1+hr*(-i2*RL)

𝑣1
By solving we get Zi= = ℎ𝑖 + 𝐴𝑖 ∗ ℎ𝑟 ∗ RL
𝑖1

ℎ𝑟 ∗ℎ𝑓∗𝑅𝐿
By substituting Ai value we will get Zi=hi-( )
1+ℎ𝑜𝑅𝐿
Expression for Voltage gain:
It is defined as the ratio of output voltage to input voltage. It is represented by Av.
Av=Vo/Vi=V2/V1=(-i2*RL)/V1 (since V2=-i2*RL)
Multiply and divide by i1 on R.H.S
−𝑖2∗𝑅𝐿
Av=( )*(i1/i1)
𝑉1

−𝑖2 𝑖1∗𝑅𝐿 𝑅𝐿
=( ) ∗ ( )=Ai*( )
𝑖1 𝑉1 Zi
ℎ𝑟 ∗ℎ𝑓∗𝑅𝐿
Where Zi=hi-( ) and Ai=-hf/(1+ho*RL)
1+ℎ𝑜𝑅𝐿

−ℎ𝑓∗𝑅𝐿
Finally, Av=( ) Where 𝜟h=hi*ho-hr*hf
ℎ𝑖+𝜟ℎ∗𝑅𝐿
Expression for Output Impedance:
To calculate Zo we need to short input source Vs and open output terminal i.e.
RL=∞ and Vs=0
𝑣2
Zo= ; Substitute i2 value from h-parameter equation then we will get one equation
𝑖2
𝑣2
Zo= −−−−− −(1)
ℎ𝑓 ∗𝑖1+ℎ𝑜∗𝑣2

Now apply KVL in input loop to get 2nd equation

-i1*Rs-i1*hi-hr*v2=0
 −ℎ𝑟∗𝑣2 𝑉2 −ℎ𝑓∗𝑅𝐿
i1= ----------------(2) Av= =( )
(𝑅𝑠+ℎ𝑖) 𝑉1 ℎ𝑖 +𝜟ℎ∗𝑅𝐿

Substitute Eq-(2) in Eq-(1)

𝑅𝑠+ℎ𝑖
Zo=
𝜟ℎ+ℎ𝑜∗𝑅𝑠

Overall voltage gain:

𝑉1
Therefore, Avs=Av* -------(3)
It is defined as ratio of output voltage to 𝑉𝑠

source voltage From the above loop

𝑉2 𝑍𝑖
Avs= V1/Vs= --------(4)
𝑉𝑠 𝑅𝑠+𝑍𝑖

Multiply and divide it by v1

𝑉2 𝑉1
Avs= *
𝑉𝑠 𝑉1
 𝑍𝑖
Therefore, Avs=Av( )
𝑅𝑠+𝑍𝑖

If Vs is ideal==>Rs=0
Therefore, Avs=Av
Overall current gain:
It is defined as the ratio of output current
Now, use current divider rule,
to the current delivered by the source. It is
𝑖𝑠∗𝑅𝑠 𝑖1 𝑅𝑠
i1= ==> = −−−−− −(5)
denoted by Ais. 𝑅𝑠+𝑍𝑖 𝑖𝑠 𝑅𝑠+𝑍𝑖

𝑅𝑠
Ais=
𝑖𝑙
∗
𝑖2
= Ai ∗
i1 Therefore, Ais=Ai*( )
𝑅𝑠+𝑍𝑖
𝑖1 𝑖𝑠 𝑖𝑠
In case of ideal current source, Rs= ∞
Convert voltage source at the input into Therefore, Ais=Ai
current source.
Approximate hybrid equivalent model of transistor:
Lets consider CE transistor

For CE and CB the magnitude of hr and ho are such that the results obtained for the
parameters like input Z, Output Z, voltage gain and current gain are slightly effected,
if not included in the circuit. Hence we can remove it from the circuit.
 1
hoe is output admittance and is output impedance which is very large compared to
ℎ𝑜𝑒

the load resistance RL. Hence we can neglect the output impedance as no chance of
flowing current through it.

==>

If transistor is connected in fixed bias configuration then there is one more resistance
1
connected in parallel to RL which is RC then also is greater than equivalent resistance
ℎ𝑜𝑒

so we can neglect in that case too.

 𝑉
hr= 𝑖 =0(approx.) (Since when transistor acts as an amplifier Vo>Vi)
𝑉𝑜

Therefore, hreVo=0 So we can replace this branch with an short circuit.

Final Equivalent Circuit:

==>

Conversion of h-parameters:
The need for the conversion is, generally the transistor manufacturer provides the h-
parameters of transistors in CE model because its mostly used.
Conversion requires following formulae:
For CECB
ℎ𝑖𝑒 ℎ𝑖𝑒∗ℎ𝑜𝑒 −ℎ𝑓𝑒 ℎ𝑜𝑒
hib= ; hrb= -hre ; hfb= ; hob=
1+ℎ𝑓𝑒 1+ℎ𝑓𝑒 1+ℎ𝑓𝑒 1+ℎ𝑓𝑒

For CECC
hic=hie ; hrc=1-hre ; hfe=-(1+hfe) ; hoc=hoe
For CBCC
ℎ𝑖𝑏 ℎ𝑖𝑏∗ℎ𝑜𝑏 −ℎ𝑓𝑒 ℎ𝑜𝑏
hie= ; hre= -h ; hfe= ; hoe=
1+ℎ𝑓𝑏 1+ℎ𝑓𝑏 rb 1+ℎ𝑓𝑏 1+ℎ𝑓𝑏
 re Transistor model
Lets find out re model for CE transistor


• Modify the above circuit by placing a dependent current source in the collector
branch and the emitter branch will have forward biased diode.

• We have dependent current source because current ic=ᵝ*ib + (ᵝ+1)*ICBO

Here we neglected the reverse saturation current as it is very small.
From the output characteristics of CE transistor, we find that the output resistance is
1
very large because resistance (ro)= (Since slope =0(approx.))
𝑠𝑙𝑜𝑝𝑒

ro=∞
In the collector branch place a forward biased diode because the input characteristics
of CE transistor is similar to the input characteristics of a forward biased diode.
There are three types of diode resistances:
i)DC resistance
ii)AC resistance(dynamic resistance)
iii)Average AC resistance
For AC analysis we will consider second type of diode resistance which is represented
𝜟𝑉𝐵𝐸
as rd. Where rd=
𝜟𝐼𝐵

Now again replace diode by dynamic resistance re(re=rd) where ie is the current flows
through it.
 ≡

Calculating re:
Consider a PN junction diode
Where ID is diode current and IS be the reverse saturation current.
𝑉
𝐷
From diode current equation we know that ID=IS(e η𝑉
𝑇
− 1)
Differentiate with respect to VD
𝑉
𝑑𝐼𝐷 𝑑 𝐷
= IS* (e
η𝑉 − 1)
𝑑𝑉𝐷 𝑑𝑉𝐷 𝑇

For high diode currents η=1

𝑑𝐼𝐷 1
And = ; Substituting these values and by the above equation we will get
𝑑𝑉𝐷 𝑟𝑑
𝑉

1 IS (e )𝐷
η𝑉
𝑇 ID+IS 𝐼𝐷
= = =
𝑟𝑑 𝑉𝑇 𝑉𝑇 𝑉𝑇

𝑉𝑇 26𝑚𝑉
Therefore, 𝑟𝑑 = = (Since at room temperature VT=26mV)
𝐼𝐷 𝐼𝐷
26𝑚𝑉
re =
𝐼𝐸
re model for CE transistor further modification:
CE is mostly used because amplification is large in this configuration.
But in the above circuit for CE model input and output sides are not properly
separated.
We know that ie = ib+ic
Substitute ic = ẞib then we will get ie = ib(1+ẞ)
Find drop across resistance re
Drop across re = re*ie = re*ib(1+ẞ)
To separate input and output circuit we will take re and (1+ẞ) together and ib is the
current flowing through the resistance re(1+ẞ)
And at the output side we have a dependent current source and a resistor ro
Therefore the final equivalent re model for CE configuration is

And we can further simplify it by placing ẞ in place of (1+ ẞ)

re model for Common Base Transistor:

In this CB, the output characteristics have 0 slope

1 1
ro = = =∞
𝑠𝑙𝑜𝑝𝑒 0

So we neglected the output resistance in re model of CB transistor.

 Hybrid-∏ model
• Widely used because we can use it for high frequency small signals and we can
also use it for low frequency small signals.
• At low frequencies it is assumed that transistor responds instantaneously to
changes in the input voltage or current.
• If frequency of the input is high (MHz) and the amplitude of the input signal is
changing the Transistor amplifier will not be able to respond. It is because; the
carriers from the emitter side will have to be injected into the collector side.
These take definite amount of time to travel from Emitter to Base, however small
it may be.
• But if the input signal is varying at much higher speed than the actual time
taken by the carries to respond, then the Transistor amplifier will not respond
instantaneously. Thus, the junction capacitances of the transistor, puts a limit to
the highest frequency signal which the transistor can handle.

≡
At high frequencies, parameters like junction capacitances come into picture.
Cu – This is the capacitor which represent Early effect and it is of few Pico Farads.
Cᴨ - Diffusion capacitance that represent minority carrier storage in base region and its
value lies between 1PF to 2PF.
rb – This is very small value which represents the resistance due to base connection
and other resistances like base spreading resistance is also included. As it is
small we can replace it by a short circuit.
rᴨ - input resistance between base and emitter terminal.
where rᴨ = ẞre
ru - resistance between base and collector terminal. It is very large, hence replace it
with open circuit.
Parameter calculations at low frequencies:
Input Conductance (gb'e):
At low frequencies, capacitive reactance will be very large and can be considered as
Open circuit. So in the hybrid-π equivalent circuit which is valid at low frequencies,
all the capacitances can be neglected. B’ =internal node in base.
Base Spreading Resistance (rb or rbb’):
The input resistance with the output shorted is hie. If output is shorted, i.e., Collector and
Emitter are joined; rb'e is in parallel with rb’c.
Output Conductance (gce)
This is the conductance with input open circuited. In h-parameters it is represented as
hoe. For Ib= 0, we have,
Validity of hybrid-π model The high frequency hybrid Pi or Giacoletto model of BJT is
valid for frequencies less than the unit gain frequency.
Trans-conductance or Mutual Conductance (gm):

The transconductance is the ratio of change in the collector current due to small changes in
the voltage VBE across the emitter junction. It is given as

-----------------(1)
We know that, the collector current in active region is given as

Substitute ∂Ic in Eq-(1) we get,

Vb’e =VE
 Problems
1) A good transconductance amplifier should have
(A) high input resistance and low output resistance GATE(2017)
(B) low input resistance and high output resistance
(C) high input and output resistances
(D) low input and output resistances
Solution: (C )
For a transconductance amplifier, input and output resistance is high.
Reason: The transconductance amplifier is also known as Voltage Controlled Current
Source. An amplifier is VC when input resistance is high, and an amplifier is CS when
output resistance is high.
2) A bipolar transistor is operating in the active region with a collector current of 1mA. Assuming that the ẞ
of the transistor is 100 and the thermal voltage(VT) is 25mV, the transconductance (gm) and the input
resistance (rᴨ) of the transistor in the common emitter configuration, are
GATE(2004)

A)gm=25mA/V and rᴨ = 15.625KΩ B)gm=40mA/V and rᴨ = 4.0KΩ

C)gm=25mA/V and rᴨ = 2.5KΩ D)gm=40mA/V and rᴨ = 2.5KΩ
Solution:
Given,
Ic = 1mA; ẞ = 100; VT = 25mV
𝐼𝐶 1 40𝑚𝐴
We know that gm = = = /v
𝑉𝑇 25

ẞ 100
ẞ=gmrᴨ → rᴨ = = − = 2.5𝑘Ω
𝑔𝑚 40∗10 3
 𝐾𝑇
3) A BJT is biased in forward active mode. Assume VBE = 0.7 V, = 25mV and reverse
𝑞

saturation current Is = 10(-3) A. The transconductance of the BJT (in mA/V) is ___
A)1.425A/V B)5784mA/V GATE(2014)
C)5790mA/V D)2675mA/V
Solution: (B)
Given,
𝐾𝑇
VBE = 0.7V; VT = = 25mV; Is = 10(-3) A
𝑞
𝑉
𝐷
We know that, IC=IS(eη𝑉𝑇 − 1) where η = 1 when diode current is high

By substituting we will get IC = 144.6mA

𝐼𝐶 144.6𝑚𝐴
Therefore, gm = = = 5784mA/V
𝑉𝑇 25m𝑉
4) The input impedance(Zi) and the output impedance(Zo) of an ideal transconductance
amplifier is ___________ GATE(2006)

A) Zi = 0, Zo = 0 B) Zi = 0, Zo =∞
C) Zi = ∞ , Zo = 0 D) Zi = ∞ , Zo =∞
Solution:
For Transconductance amplifier Zi = ∞ , Zo = ∞
5) The current ib through base of a silicon npn transistor is 1+0.1 cos (1000πt) ma. At
300K, the rπ in the small signal model of the transistor is GATE-2012

(a) 250Ω (b) 27.5Ω (c) 25Ω (d) 22.5Ω

Solution: (C).
Current ib through the base of a silicon npn transistor is 1+0.1 cos (10000 πt) ma
rπ= 𝜷. 𝒓𝒆 = 𝜷 𝑽𝑻 /IE ≅ 𝜷 𝑽𝑻 /𝜷𝒊𝒃 = 𝑽𝑻 /𝒊b
𝑽𝑻 = 𝟐𝟓𝒎𝒗,𝒊𝒃 = 𝟏𝒎𝒂
𝒓𝝅 = 𝟐𝟓 Ω
6) The current gain of a BJT is GATE-2002
(a) gmr0 (b) gm / ro (c) gmrπ (d) gm / rπ

Solution:

𝒈𝒎 = 𝑰𝑪 / VT = 𝜷 𝑰𝑩 /𝑰𝑩𝒓𝝅 𝒈𝒎 = 𝜷/ 𝒓𝝅
𝒔𝒐 𝜷 = 𝒈𝒎𝒓𝝅